How do you find the vertical, horizontal or slant asymptotes for #y=(2x^2-3x+4)/(x+2)#?

1 Answer
Jun 3, 2016

Answer:

Vertical asymptotes at #x_v = -2#
Slant asymptote #y = 2x-7#

Explanation:

In a polynomial fraction #f(x) = (p_n(x))/(p_m(x))# we have:

#1)# vertical asymptotes for #x_v# such that #p_m(x_v)=0#
#2)# horizontal asymptotes when #n le m#
#3)# slant asymptotes when #n = m + 1#
In the present case we have #x_v = -2# and #n = m+1# with #n = 2# and #m = 1#

Slant asymptotes are obtained considering #(p_n(x))/(p_{n-1}(x)) approx y = a+b x# for large values of #abs(x)#

In the present case we have

#(p_n(x))/(p_{n-1}(x)) = (2x^2-3x+4)/(x + 2) #
#p_n(x) = p_{n-1}(x)(a x + b) + r_{n-2}(x)#
#(p_n(x))/(p_{n-1}(x)) = (2x^2-3x+4) =(x+2) (a x +b)+c#

equating we have

#{ (4 - 2 b - c=0), (-3 - 2 a - b=0), (2 - a=0) :}#

Solving for #a,b,c# we have #a = 2,b=-7,c=18#
so the slant asymptote reads

#y = 2x-7 #

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